CAP7001 · Unit 2: Physical layer
Analog and digital signals
11 practice questions · in the MTE and ETE syllabus
Key points
- Analog = continuous; digital = discrete levels
- f = 1/T; bandwidth = f_max - f_min
- Bit rate = baud x log2(L)
- Baud = bit rate / bits per element
An analog signal has infinitely many values over time; a digital signal has a limited set of discrete levels. Periodic signals are described by amplitude, frequency and phase, where f = 1/T (T is the period). The bandwidth of a composite signal is f_max - f_min.
For digital signals:
- Bit rate = bits per second.
- Baud rate (signal rate) = signal elements per second.
- Bit rate = baud rate x log2(L) for L signal levels, so baud = bit rate / bits per element.
Worked example: 6000 bps with 3 bits per element gives 2000 baud. A line fixed at 2400 baud moving from 4 levels (2 bits) to 64 levels (6 bits) triples its bit rate.
| System | Levels | Baud | Bit rate |
|---|---|---|---|
| A | 2 | 9600 | 9600 bps |
| B | 16 | 2400 | 9600 bps |
Exam pitfall: bit rate and baud are different. More levels raise the bit rate only with log2(L), so 64 levels is 3 times, not 16 times, the rate of 4 levels.
Try two questions
The period of a periodic signal is doubled. What happens to its frequency?
- It is doubled
- It stays the same
- It is halved
- It is squared
Show the answer
C. Frequency is the reciprocal of the period, f = 1/T, so doubling T halves f.
Why not A: Frequency moves in the opposite direction to the period.
Why not B: The two quantities are linked, so a change in period changes frequency.
Why not D: A reciprocal relation does not square anything.
A noiseless channel sends 4-level signals at 9600 bps. Using the Nyquist formula, what is the minimum bandwidth of the channel?
- 2400 Hz
- 4800 Hz
- 1200 Hz
- 9600 Hz
Show the answer
A. With 4 levels each element carries 2 bits, and bit rate = 2 x B x log2(4) = 4B, so B = 9600 / 4 = 2400 Hz. The 4800 Hz figure forgets the log2(4) factor.
Why not B: That ignores the 2 bits per element and uses bit rate = 2B.
Why not C: That divides once more than needed, using 8 as the divisor.
Why not D: That treats the bit rate itself as the bandwidth.